2023 AMC 12A Problems/Problem 8

The following problem is from both the 2023 AMC 10A #10 and 2023 AMC 12A #8, so both problems redirect to this page.


Maureen is keeping track of the mean of her quiz scores this semester. If Maureen scores an $11$ on the next quiz, her mean will increase by $1$. If she scores an $11$ on each of the next three quizzes, her mean will increase by $2$. What is the mean of her quiz scores currently? $\textbf{(A) }4\qquad\textbf{(B) }5\qquad\textbf{(C) }6\qquad\textbf{(D) }7\qquad\textbf{(E) }8$

Solution 1

Let $a$ represent the amount of tests taken previously and $x$ the mean of the scores taken previously.

We can write the following equations:

\[\frac{ax+11}{a+1}=x+1\qquad (1)\] \[\frac{ax+33}{a+3}=x+2\qquad (2)\]

Multiplying $(x+1)$ by $(a+1)$ and solving, we get: \[ax+11=ax+a+x+1\] \[11=a+x+1\] \[a+x=10\qquad (3)\]

Multiplying $(2)$ by $(a+3)$ and solving, we get: \[ax+33=ax+2a+3x+6\] \[33=2a+3x+6\] \[2a+3x=27\qquad (4)\]

Solving the system of equations for $(3)$ and $(4)$, we find that $a=3$ and $x=\boxed{\textbf{(D) }7}$.

~walmartbrian ~Shontai ~andyluo ~megaboy6679

Solution 2 (Variation on Solution 1)

Suppose Maureen took $n$ tests with an average of $m$.

If she takes another test, her new average is $\frac{(nm+11)}{(n+1)}=m+1$

Cross-multiplying: $nm+11=nm+n+m+1$, so $n+m=10$.

If she takes $3$ more tests, her new average is $\frac{(nm+33)}{(n+3)}=m+2$

Cross-multiplying: $nm+33=nm+2n+3m+6$, so $2n+3m=27$.

But $2n+3m$ can also be written as $2(n+m)+m=20+m$. Therefore $m=27-20=\boxed{\textbf{(D) }7}$

~Dilip ~megaboy6679 (latex)

Solution 3 (do this if you are bored)

Let $s$ represent the sum of Maureen's test scores previously and $t$ be the number of scores taken previously.

So, $\frac{s+11}{t+1} = \frac{s}{t}+1$ and $\frac{s+33}{t+3} = \frac{s}{t}+2$

We can use the first equation to write $s$ in terms of $t$.

We then substitute this into the second equation: $\frac{-t^2+10t+33}{t+3} = \frac{-t^2+10t}{t}+2$

From here, we solve for t: multiply both sides by ($t$) and then ($t+3$), combining like terms to get $t^2-3t=0$. Factorize to get $t=0$ or $t=3$, and therefore $t=3$ (makes sense for the problem).

We substitute this to get $s=21$.

Therefore, the solution to the problem is $\frac{21}{3}=$ $\boxed{\textbf{(D) }7}$

~milquetoast ~the_eaglercraft_grinder

Solution 4 (Trial and Error)

Let's consider all the answer choices. If the average is $8$, then, we can assume that all her test choices were $8$. We can see that she must have gotten $8$ twice, in order for another score of $11$ to bring her average up by one. However, adding three $11$'s will not bring her score up to 10. Continuing this process for the answer choices, we see that the answer is $\boxed{\textbf{(D) }7}$ ~andliu766

Solution 5

Let $n$ be the number of existing quizzes. So after one more test, score $11$ has $n+1$ extra points to distribute to $n+1$ quizzes. Also, after three more quizzes, there will be $3(n+1)$ extra points to distribute to the $n+3$ quizzes. So $3n+3=2(n+3)$. This means $n=3$. $n+1$ extra points means original mean (average) is $7$


Video Solution by Power Solve (easy to digest!)


Video Solution (🚀 Just 3 min 🚀)


~Education, the Study of Everything

Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/cMgngeSmFCY?si=MHL95YihFdxKROrU&t=2280 ~Math-X

Video Solution by CosineMethod [🔥Fast and Easy🔥]


Video Solution


~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

See Also

2023 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions
2023 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png